1. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Introductory Chemistry, 2 nd Edition Nivaldo Tro Chapter 13 Solutions 2006, Prentice Hall

2. Tro's Introductory Chemistry, Chapter 13 2 Tragedy in Cameroon Lake Nyos lake in Cameroon, West Africa on August 22, 1986, 1700 people & 3000 cattle died Burped Carbon Dioxide Cloud CO 2 seeps in from underground and dissolves in lake water to levels above normal saturation though not toxic, CO 2 is heavier than air – the people died from asphyxiation

3. Tro's Introductory Chemistry, Chapter 13 3 Tragedy in Cameroon: a Possible Solution scientist have studied Lake Nyos and similar lakes in the region to try and keep such a tragedy from reoccurring currently, they are trying to keep the CO 2 levels in the lake water from reaching the very high supersaturation levels by pumping air into the water to agitate it

4. Tro's Introductory Chemistry, Chapter 13 4 Solution homogeneous mixtures composition may vary from one sample to another appears to be one substance, though really contains multiple materials most homogeneous materials we encounter are actually solutions e.g. air and lake water

5. Tro's Introductory Chemistry, Chapter 13 5 Solutions solute is the dissolved substance seems to “disappear” “takes on the state” of the solvent solvent is the substance solute dissolves in does not appear to change state when both solute and solvent have the same state, the solvent is the component present in the highest percentage solutions in which the solvent is water are called aqueous solutions

6. 6 Brass TypeColor% Cu% ZnDensity g/cm 3 MP °C Tensile Strength psi Uses Gildingreddish9558.86106650Kpre-83 pennies, munitions, plaques Commercialbronze90108.80104361Kdoor knobs, grillwork Jewelrybronze87.512.58.78103566Kcostume jewelry Redgolden85158.75102770Kelectrical sockets, fasteners & eyelets Lowdeep yellow 80208.6799974Kmusical instruments, clock dials Cartridgeyellow70308.4795476Kcar radiator cores Commonyellow67338.4294070Klamp fixtures, bead chain Muntz metalyellow60408.3990470Knuts & bolts, brazing rods

7. Tro's Introductory Chemistry, Chapter 13 7 Common Types of Solution Solution Phase Solute Phase Solvent PhaseExample gaseous solutionsgas air (mostly N 2 & O 2 ) liquid solutions gas liquid solid liquid soda (CO 2 in H 2 O) vodka (C 2 H 5 OH in H 2 O) seawater (NaCl in H 2 O) solid solutionssolid brass (Zn in Cu)

8. Tro's Introductory Chemistry, Chapter 13 8 Solubility solutions that contain Hg and some other metal are called amalgams solutions that contain metal solutes and a metal solvent are called alloys when one substance (solute) dissolves in another (solvent) it is said to be soluble salt is soluble in water, bromine is soluble in methylene chloride when one substance does not dissolve in another it is said to be insoluble oil is insoluble in water

9. Tro's Introductory Chemistry, Chapter 13 9 Will It Dissolve? Chemist’s Rule of Thumb – Like Dissolves Like a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other

10. 10 Classifying Solvents SolventClass Structural Feature Water, H 2 OpolarO-H Ethyl Alcohol, C 2 H 5 OHpolarO-H Acetone, C 3 H 6 OpolarC=O Benzene, C 6 H 6 nonpolarC-C & C-H Hexane, C 6 H 14 nonpolarC-C & C-H Diethyl Ether, C 4 H 10 OnonpolarC-C, C-H & C-O

11. Tro's Introductory Chemistry, Chapter 13 11 Will It Dissolve In Water? ions are attracted to polar solvents many ionic compounds dissolve in water polar molecules are attracted to polar solvents table sugar, ethyl alcohol and glucose all dissolve well in water nonpolar molecules are attracted to nonpolar solvents  -carotene, (C 40 H 56 ), is not water soluble; it dissolves in fatty (nonpolar) tissues many molecules have both polar and nonpolar structures – whether they will dissolve in water depends on the kind, number and location of polar and nonpolar structural features in the molecule

12. Tro's Introductory Chemistry, Chapter 13 12 Salt Dissolving in Water

13. Tro's Introductory Chemistry, Chapter 13 13 Solvated Ions When materials dissolve, the solvent molecules surround the solvent particles due to the solvent’s attractions for the solute. The process is called solvation. Solvated ions are effectively isolated from each other.

14. Tro's Introductory Chemistry, Chapter 13 14 Solubility there is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible  alcohol and water are miscible  oil and water are immiscible the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility

15. Tro's Introductory Chemistry, Chapter 13 15 Descriptions of Solubility saturated solutions have the maximum amount of solute that will dissolve in that solvent at that temperature unsaturated solutions can dissolve more solute supersaturated solutions are holding more solute than they should be able to at that temperature unstable

16. Tro's Introductory Chemistry, Chapter 13 16 Supersaturated Solution A supersaturated solution has more dissolved solute than the solvent can hold. When disturbed, all the solute above the saturation level comes out of solution.

17. Tro's Introductory Chemistry, Chapter 13 17 Adding Solute to various Solutions unsaturated saturated supersaturated

18. Tro's Introductory Chemistry, Chapter 13 18 Electrolytes electrolytes are substances whose aqueous solution is a conductor of electricity in strong electrolytes, all the electrolyte molecules are dissociated into ionsstrong electrolytes in nonelectrolytes, none of the molecules are dissociated into ions in weak electrolytes, a small percentage of the molecules are dissociated into ions

19. 19 Solubility and Temperature the solubility of the solute in the solvent depends on the temperature higher temp = higher solubility of solid in liquid lower temp = higher solubility of gas in liquid

20. Tro's Introductory Chemistry, Chapter 13 20 Solubility and Temperature Warm soda pop fizzes more than cold soda pop because the solubility of CO 2 in water decreases as temperature increases.

21. Tro's Introductory Chemistry, Chapter 13 21 Solubility and Pressure the solubility of gases in water depends on the pressure of the gas higher pressure = higher solubility

22. Tro's Introductory Chemistry, Chapter 13 22 Solubility and Pressure When soda pop is sealed, the CO 2 is under pressure. Opening the container lowers the pressure, which decreases the solubility of CO 2 and causes bubbles to form.

23. Solution Concentrations

24. Tro's Introductory Chemistry, Chapter 13 24 Solution Concentration Descriptions dilute solutions have low solute concentrations concentrated solutions have high solute concentrations

25. Tro's Introductory Chemistry, Chapter 13 25 Concentrations – Quantitative Descriptions of Solutions Solutions have variable composition To describe a solution accurately, you need to describe the components and their relative amounts Concentration = amount of solute in a given amount of solution Occasionally amount of solvent

26. Tro's Introductory Chemistry, Chapter 13 26 Mass Percent parts of solute in every 100 parts solution if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution  or 10 kg solute in every 100 kg solution since masses are additive, the mass of the solution is the sum of the masses of solute and solvent

27. Example 13.1: Calculating Mass Percent

28. Tro's Introductory Chemistry, Chapter 13 28 Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL)

29. Tro's Introductory Chemistry, Chapter 13 29 Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL) Write down the given quantity and its units. Given:27.5 g C 2 H 6 O 175 mL H 2 O

30. Tro's Introductory Chemistry, Chapter 13 30 Write down the quantity to find and/or its units. Find: mass percent Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL)

31. Tro's Introductory Chemistry, Chapter 13 31 Collect Needed Equations: Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL)

32. Tro's Introductory Chemistry, Chapter 13 32 Collect Needed Conversion Factors: Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Eq’n: Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL) d(H 2 O) = 1.00 g/mL  1.00 g H 2 O = 1 mL H 2 O

33. Tro's Introductory Chemistry, Chapter 13 33 Design a Solution Map: Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Eq’n: CF:1.00 g H 2 O = 1 mL H 2 O Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL) Mass Solution Mass Solute & Mass PercentVol Solvent Mass Solvent density Solution = solute + solvent

34. Tro's Introductory Chemistry, Chapter 13 34 Apply the Solution Maps Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Eq’n: CF:1.00 g H 2 O = 1 mL H 2 O SM: mass sol & vol solv → mass solv → mass sol’n → mass percent Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL) Mass of Solution = Mass C 2 H 6 O + Mass H 2 O = 27.5 g C 2 H 6 O + 175 g H 2 O = 202.5 g

35. 35 Apply the Solution Maps - Equation = 13.5802% Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Eq’n: CF:1.00 g H 2 O = 1 mL H 2 O SM: mass sol & vol solv → mass solv → mass sol’n → mass percent Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL) = 13.6%

36. 36 Check the Solution The units of the answer, %, are correct. The magnitude of the answer makes sense since the mass of solute is less than the mass of solvent. Mass Percent = 13.6% Information Given:27.5 g C 2 H 6 O; 175 mL H 2 O Find:% by mass Eq’n: CF:1.00 g H 2 O = 1 mL H 2 O SM: mass sol & vol solv → mass solv → mass sol’n → mass percent Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O (assume the density of H 2 O is 1.00 g/mL)

37. Tro's Introductory Chemistry, Chapter 13 37 Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent 12% by mass sugar(aq) means 12 g sugar  100 g solution The concentration can then be used to convert the amount of solute into the amount of solution, or visa versa

38. Example 13.2: Using Mass Percent in Calculations

39. Tro's Introductory Chemistry, Chapter 13 39 Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)

40. Tro's Introductory Chemistry, Chapter 13 40 Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) Write down the given quantity and its units. Given:85.2 g C 12 H 22 O 11

41. Tro's Introductory Chemistry, Chapter 13 41 Write down the quantity to find and/or its units. Find: volume of solution, mL Information Given:85.2 g C 12 H 22 O 11 Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)

42. Tro's Introductory Chemistry, Chapter 13 42 Collect Needed Conversion Factors: d = 1.00 g/mL  1.00 g solution = 1 mL solution Information Given:85.2 g C 12 H 22 O 11 Find:mL sol’n Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) 11.5% by mass  11.5 g C 12 H 22 O 11  100 g solution

43. Tro's Introductory Chemistry, Chapter 13 43 Design a Solution Map: Mass Solution Mass SoluteVolume Solution density Information Given:85.2 g C 12 H 22 O 11 Find:mL sol’n CF:11.5 g C 12 H 22 O 11  100 g sol’n 1.00 g sol’n = 1 mL sol’n Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) mass percent

44. Tro's Introductory Chemistry, Chapter 13 44 Apply the Solution Map Information Given:85.2 g C 12 H 22 O 11 Find:mL sol’n CF:11.5 g C 12 H 22 O 11  100 g sol’n 1.00 g sol’n = 1 mL sol’n SM: g sucrose → g sol’n → mL sol’n Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) = 740.87 mL = 741 mL

45. 45 Check the Solution The units of the answer, mL, are correct. The magnitude of the answer makes sense since the mass of solute is less than the volume of solution. volume of solution = 741 mL Information Given:85.2 g C 12 H 22 O 11 Find:mL sol’n CF:11.5 g C 12 H 22 O 11  100 g sol’n 1.00 g sol’n = 1 mL sol’n SM: g sucrose → g sol’n → mL sol’n Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)

46. Tro's Introductory Chemistry, Chapter 13 46 Preparing a Solution need to know amount of solution and concentration of solution calculate the mass of solute needed start with amount of solution use concentration as a conversion factor 5% by mass solute  5 g solute  100 g solution Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)? dissolve 12.5 g of glucose in enough water to total 250 g

47. Tro's Introductory Chemistry, Chapter 13 47 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar molarity = moles of solute liters of solution

48. Tro's Introductory Chemistry, Chapter 13 48 Preparing a 1.00 M NaCl Solution Weigh out 1 mole (58.45 g) of NaCl and add it to a 1.00 L volumetric flask. Step 1 Step 2 Add water to dissolve the NaCl, then add water to the mark. Step 3 Swirl to Mix

49. Example 13.3: Calculating Molarity

50. Tro's Introductory Chemistry, Chapter 13 50 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

51. Tro's Introductory Chemistry, Chapter 13 51 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. Write down the given quantity and its units. Given:15.5 g NaCl 1.50 L solution

52. Tro's Introductory Chemistry, Chapter 13 52 Write down the quantity to find and/or its units. Find: molarity (M) Information Given:15.5 g NaCl; 1.50 L sol’n Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

53. Tro's Introductory Chemistry, Chapter 13 53 Collect Needed Equations and Conversion Factors: Molar Mass NaCl = 58.44 g/mol  58.44 g NaCl = 1 mol Information Given:15.5 g NaCl; 1.50 L sol’n Find: molarity, M Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

54. 54 Design a Solution Map: Volume Solution Mass Solute Molarity Information Given:15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. L Solution Mole Solute already liters

55. Tro's Introductory Chemistry, Chapter 13 55 Apply the Solution Map Information Given:15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. = 0.177 M NaCl

56. 56 Check the Solution The units of the answer, M, are correct. The magnitude of the answer makes sense since the mass of solute is less than the 1 mole and the volume is more than 1 L. molarity of solution = 0.177 M Information Given:15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

57. Example 13.4: Using Molarity in Calculations

58. Tro's Introductory Chemistry, Chapter 13 58 Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

59. Tro's Introductory Chemistry, Chapter 13 59 Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH? Write down the given quantity and its units. Given:1.24 mol NaOH

60. Tro's Introductory Chemistry, Chapter 13 60 Write down the quantity to find and/or its units. Find: volume of solution (L) Information Given:1.24 mol NaOH Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

61. Tro's Introductory Chemistry, Chapter 13 61 Collect Needed Conversion Factors: Molarity = 0.114 mol/L  0.114 mol NaOH = 1 L sol’n Information Given:1.24 mol NaOH Find: L solution Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

62. 62 Design a Solution Map: L SolutionMole Solute Information Given:1.24 mol NaOH Find: L solution CF:0.114 mol = 1 L Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

63. Tro's Introductory Chemistry, Chapter 13 63 Apply the Solution Map Information Given:1.24 mol NaOH Find: L solution CF:0.114 mol = 1 L SM: mol → L Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

64. 64 Check the Solution The units of the answer, L, are correct. The magnitude of the answer makes sense - since 1 L only contains 0.114 moles, the volume must be more than 1 L. volume of solution = 10.9 L Information Given:1.24 mol NaOH Find: L solution CF:0.114 mol = 1 L SM: mol → L Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

65. Tro's Introductory Chemistry, Chapter 13 65 How would you prepare 250 mL of 0.20 M NaCl? 0.250 L x 0.20 moles NaCl 1 L x 58.44 g 1 mole NaCl = 2.9 g NaCl Dissolve 2.9 g of NaCl in enough water to total 250 mL Sample - Molar Solution Preparation

66. Tro's Introductory Chemistry, Chapter 13 66 Molarity and Dissociation When strong electrolytes dissolve, all the solute particles dissociate into ions By knowing the formula of the compound and the molarity of the solution, it is easy to determine the molarity of the dissociated ions simply multiply the salt concentration by the number of ions

67. Tro's Introductory Chemistry, Chapter 13 67 Molarity & Dissociation NaCl(aq) = Na + (aq) + Cl - (aq) 1 “molecule”= 1 ion + 1 ion 100 “molecules”= 100 ions + 100 ions 1 mole “molecules”= 1 mole ions + 1 mole ions 1 M NaCl “molecules”= 1 M Na + ions + 1 M Cl - ions 0.25 M NaCl= 0.25 M Na + + 0.25 M Cl -

68. Tro's Introductory Chemistry, Chapter 13 68 Molarity & Dissociation CaCl 2 (aq) = Ca 2+ (aq) + 2 Cl - (aq) 1 “molecule”= 1 ion + 2 ion 100 “molecules”= 100 ions + 200 ions 1 mole “molecules”= 1 mole ions + 2 mole ions 1 M CaCl 2 = 1 M Ca 2+ ions + 2 M Cl - ions 0.25 M CaCl 2 = 0.25 M Ca 2+ + 0.50 M Cl -

69. Tro's Introductory Chemistry, Chapter 13 69 Find the molarity of all ions in the given solutions of strong electrolytes 0.25 M MgBr 2 (aq) 0.33 M Na 2 CO 3 (aq) 0.0750 M Fe 2 (SO 4 ) 3 (aq)

70. Tro's Introductory Chemistry, Chapter 13 70 Find the molarity of all ions in the given solutions of strong electrolytes MgBr 2 (aq) → Mg 2+ (aq) + 2 Br - (aq) 0.25 M0.25 M0.50 M Na 2 CO 3 (aq) → 2 Na + (aq) + CO 3 2- (aq) 0.33 M 0.66 M 0.33 M Fe 2 (SO 4 ) 3 (aq) → 2 Fe 3+ (aq) + 3 SO 4 2- (aq) 0.0750 M 0.150 M 0.225 M

71. Tro's Introductory Chemistry, Chapter 13 71 Dilution Dilution is adding extra solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases Dilution Formula Conc start soln x Vol start soln = Conc final soln x Vol final sol Concentrations and Volumes can be most units as long as consistent

72. Tro's Introductory Chemistry, Chapter 13 72 Example – What Volume of 12.0 M KCl is needed to make 5.00 L of 1.50 M KCl Solution? Given: Initial SolutionFinal Solution Concentration12.0 M1.50 M Volume? L5.00 L Find:L of initial KCl Equation:(conc 1 )∙(vol 1 ) = (conc 2 )∙(vol 2 ) Rearrange and Apply Equation:

73. Tro's Introductory Chemistry, Chapter 13 73 Making a Solution by Dilution M 1 x V 1 = M 2 x V 2 M 1 = 12.0 M V 1 = ? L M 2 = 1.50 M V 2 = 5.00 L dilute 0.625 L of 12.0 M solution to 5.00 L

74. Tro's Introductory Chemistry, Chapter 13 74 Solution Stoichiometry we know that the balanced chemical equation tells us the relationship between moles of reactants and products in a reaction 2 H 2 (g) + O 2 (g) → 2 H 2 O(l) implies for every 2 moles of H 2 you use you need 1 mole of O 2 and will make 2 moles of H 2 O since molarity is the relationship between moles of solute and liters of solution, we can now measure the moles of a material in a reaction in solution by knowing its molarity and volume

75. Example 13.7: Solution Stoichiometry

76. Tro's Introductory Chemistry, Chapter 13 76 Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq)

77. Tro's Introductory Chemistry, Chapter 13 77 Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq) Write down the given quantity and its units. Given:0.104 L Pb(NO 3 ) 2

78. Tro's Introductory Chemistry, Chapter 13 78 Write down the quantity to find and/or its units. Find: volume of KI solution, L Information Given:0.104 L Pb(NO 3 ) 2 Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq)

79. Tro's Introductory Chemistry, Chapter 13 79 Collect Needed Conversion Factors: 0.115 M KI  0.115 mol KI  1 L solution Information Given:0.104 L Pb(NO 3 ) 2 Find:L KI Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq) 0.225 M Pb(NO 3 ) 2  0.225 mol Pb(NO 3 ) 2  1 L solution Chem. Eq’n  2 mol KI  1 mol Pb(NO 3 ) 2

80. 80 Design a Solution Map: Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq) Information Given:0.104 L Pb(NO 3 ) 2 Find:L KI CF: 0.115 mol KI  1 L sol’n 0.225 mol Pb(NO 3 ) 2  1 L sol’n 2 mol KI  1 mol Pb(NO 3 ) 2 mol Pb(NO 3 ) 2 L Pb(NO 3 ) 2 mol KIL KI

81. Tro's Introductory Chemistry, Chapter 13 81 Apply the Solution Map = 0.40696 L = 0.407 L Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq) Information Given:0.104 L Pb(NO 3 ) 2 Find:L KI CF: 0.115 mol KI  1 L sol’n 0.225 mol Pb(NO 3 ) 2  1 L sol’n 2 mol KI  1 mol Pb(NO 3 ) 2 SM: L Pb(NO 3 ) 2 → mol Pb(NO 3 ) 2 → mol KI → L KI

82. Tro's Introductory Chemistry, Chapter 13 82 Check the Solution The units of the answer, L KI solution, are correct. The magnitude of the answer makes sense since the molarity of Pb(NO 3 ) 2 is larger than KI, and it takes 2x as many moles of KI as Pb(NO 3 ) 2, the volume of KI solution should be larger than the volume of Pb(NO 3)2. volume of KI solution required = 0.407 L Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb 2+ in 0.104 L of 0.225 M Pb(NO 3 ) 2 ? 2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq) Information Given:0.104 L Pb(NO 3 ) 2 Find:L KI CF: 0.115 mol KI  1 L sol’n 0.225 mol Pb(NO 3 ) 2  1 L sol’n 2 mol KI  1 mol Pb(NO 3 ) 2 SM: L Pb(NO 3 ) 2 → mol Pb(NO 3 ) 2 → mol KI → L KI

83. Tro's Introductory Chemistry, Chapter 13 83 Why do we do that? we spread salt on icy roads and walkways to melt the ice we add antifreeze to car radiators to prevent the water from boiling or freezing antifreeze is mainly ethylene glycol when we add solutes to water, it changes the freezing point and boiling point of the water

84. Tro's Introductory Chemistry, Chapter 13 84 Colligative Properties the properties of the solution are different from the properties of the solvent any property of a solution whose value depends only on the number of dissolved solute particles is called a colligative property it does not depend on what the solute particle is the freezing point, boiling point and osmotic pressure of a solution are colligative properties

85. Tro's Introductory Chemistry, Chapter 13 85 Solution Concentration Molality, m moles of solute per 1 kilogram of solvent defined in terms of amount of solvent, not solution does not vary with temperature  because based on masses, not volumes mass of solution = mass of solute + mass of solvent mass of solution = volume of solution x density

86. Example 13.8: Calculating Molality

87. Tro's Introductory Chemistry, Chapter 13 87 Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

88. Tro's Introductory Chemistry, Chapter 13 88 Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water. Write down the given quantity and its units. Given:17.2 g C 2 H 6 O 2 0.500 kg H 2 O

89. Tro's Introductory Chemistry, Chapter 13 89 Write down the quantity to find and/or its units. Find: molality (m) Information Given:17.2 g C 2 H 6 O 2 ; 0.500 kg H 2 O Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

90. Tro's Introductory Chemistry, Chapter 13 90 Collect Needed Equations and Conversion Factors: Molar Mass C 2 H 6 O 2 = 62.08 g/mol  62.08 g C 2 H 6 O 2 = 1 mol Information Given:17.2 g C 2 H 6 O 2 ; 0.500 kg H 2 O Find: molality, m Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

91. 91 Design a Solution Map: Mass Solvent Mass Solute Molality kg Solvent Mole Solute already kg Information Given:17.2 g C 2 H 6 O 2 ; 0.500 kg H 2 O Find: molality, m CF:62.08 g C 2 H 6 O 2 = 1 mol Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

92. Tro's Introductory Chemistry, Chapter 13 92 Apply the Solution Map = 0.554 m C 2 H 6 O 2 Information Given:17.2 g C 2 H 6 O 2 ; 0.500 kg H 2 O Find: molality, m CF:62.08 g C 2 H 6 O 2 = 1 mol Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

93. 93 Check the Solution The units of the answer, m, are correct. The magnitude of the answer makes sense since the mass of solute is less than the ½ mole and the mass of solvent is ½ kg. molality of solution = 0.554 m Information Given:17.2 g C 2 H 6 O 2 ; 0.500 kg H 2 O Find: molality, m CF:62.08 g C 2 H 6 O 2 = 1 mol Example: Calculate the molality of a solution containing 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 0.500 kg of water.

94. Tro's Introductory Chemistry, Chapter 13 94 Freezing Points of Solutions the freezing point of a solution is always lower than the freezing point of a pure solvent Freezing Point Depression the difference between the freezing points of the solution and pure solvent is directly proportional to the molal concentration  T f = m x K f K f = freezing point constant used to determined molar mass of compounds

95. Freezing & Boiling Point Constants

96. Example 13.9: Freezing Point Depression

97. Tro's Introductory Chemistry, Chapter 13 97 Example: Calculate the freezing point of a 1.7 m ethylene glycol solution.

98. Tro's Introductory Chemistry, Chapter 13 98 Example: Calculate the freezing point of a 1.7 m ethylene glycol solution. Write down the given quantity and its units. Given:1.7 m C 2 H 6 O 2

99. Tro's Introductory Chemistry, Chapter 13 99 Write down the quantity to find and/or its units. Find: freezing point (°C) Information Given:1.7 m C 2 H 6 O 2 in H 2 O Example: Calculate the freezing point of a 1.7 m ethylene glycol solution.

100. Tro's Introductory Chemistry, Chapter 13 100 Collect Needed Equations: Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:FP (°C) Example: Calculate the freezing point of a 1.7 m ethylene glycol solution.

101. 101 Design a Solution Map: Molality Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:FP (°C) Eq’n:  T f = m ∙ K f ; FP solution = FP solvent -  T f Example: Calculate the freezing point of a 1.7 m ethylene glycol solution. TfTf FP solution  T f = m ∙ K f FP solution  = FP solvent -  T f

102. Tro's Introductory Chemistry, Chapter 13 102 Apply the Solution Map Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:FP (°C) Eq’n:  T f = m ∙ K f ; FP solution = FP solvent –  T f SM:m →  T f → FP solution Example: Calculate the freezing point of a 1.7 m ethylene glycol solution.

103. 103 Check the Solution The units of the answer, °C, are correct. The magnitude of the answer makes sense since the freezing point of the solution is less than the freezing point of H 2 O. freezing point of solution = -3.2 °C Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:FP (°C) Eq’n:  T f = m ∙ K f ;  T f = FP solvent – FP solution SM:m →  T f → FP solution Example: Calculate the freezing point of a 1.7 m ethylene glycol solution.

104. Tro's Introductory Chemistry, Chapter 13 104 Boiling Points of Solutions the boiling point of a solution is always higher than the boiling point of a pure solvent Boiling Point Elevation the difference between the boiling points of the solution and pure solvent is directly proportional to the molal concentration  T b = m x K b K b = boiling point constant

105. Example 13.10: Boiling Point Elevation

106. Tro's Introductory Chemistry, Chapter 13 106 Example: Calculate the boiling point of a 1.7 m ethylene glycol solution.

107. Tro's Introductory Chemistry, Chapter 13 107 Example: Calculate the boiling point of a 1.7 m ethylene glycol solution. Write down the given quantity and its units. Given:1.7 m C 2 H 6 O 2

108. Tro's Introductory Chemistry, Chapter 13 108 Write down the quantity to find and/or its units. Find: boiling point (°C) Information Given:1.7 m C 2 H 6 O 2 in H 2 O Example: Calculate the boiling point of a 1.7 m ethylene glycol solution.

109. Tro's Introductory Chemistry, Chapter 13 109 Collect Needed Equations: Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:BP (°C) Example: Calculate the boiling point of a 1.7 m ethylene glycol solution.

110. 110 Design a Solution Map: Molality Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:BP (°C) Eq’n:  T b = m ∙ K b ; BP solution = BP solvent +  T b Example: Calculate the boiling point of a 1.7 m ethylene glycol solution. TbTb BP solution  T b = m ∙ K b BP solution = BP solvent +  T b

111. Tro's Introductory Chemistry, Chapter 13 111 Apply the Solution Map Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:BP (°C) Eq’n:  T b = m ∙ K b ; BP solution = BP solvent +  T b SM:m →  T b → BP solution Example: Calculate the boiling point of a 1.7 m ethylene glycol solution.

112. 112 Check the Solution The units of the answer, °C, are correct. The magnitude of the answer makes sense since the boiling point of the solution is higher than the boiling point of H 2 O. boiling point of solution = 100.87 °C Information Given:1.7 m C 2 H 6 O 2 in H 2 O Find:BP (°C) Eq’n:  T b = m ∙ K b ; BP solution = BP solvent +  T b SM:m →  T b → BP solution Example: Calculate the boiling point of a 1.7 m ethylene glycol solution.

113. Tro's Introductory Chemistry, Chapter 13 113 Osmosis & Osmotic Pressure osmosis is the process in which solvent molecules pass through a semi-permeable membrane that does not allow solute particles to pass solvent flows to try to equalize concentration of solute on both sides solvent flows from side of low concentration to high concentration osmotic pressure is pressure that is needed to prevent osmotic flow of solvent isotonic, hypotonic and hypertonic solutions hemolysis

114. Tro's Introductory Chemistry, Chapter 13 114 Drinking Seawater Because seawater has a higher salt concentration than your cells, water flows out of your cells into the seawater to try to decrease its salt concentration. The net result is that, instead of quenching your thirst, you become dehydrated.

115. Tro's Introductory Chemistry, Chapter 13 115 Osmotic Pressure Solvent flows through a semipermeable membrane to make the solution concentration equal on both sides of the membrane. The pressure required to stop this process is the osmotic pressure.

116. Tro's Introductory Chemistry, Chapter 13 116 Hemolysis & Crenation normal red blood cell in an isotonic solution red blood cell in hypotonic solution – water flows into the cell – eventually causing the cell to burst red blood cell in hypertonic solution – water flows out of the cell – eventually causing the cell to distort and shrink